anonymous
  • anonymous
Solve the non-linear inequality. Express the solution in interval notation. x^4 is greater than x^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[x^4>x^2\]
anonymous
  • anonymous
I take the square root of both sides and get the plus or minus square root of x^2 is greater than x, then I take the square root of x^2 and x to get x is greater than the plus or minus square root of x
anonymous
  • anonymous
\[x ^{4}>x ^{2}\] \[x ^{4}-x ^{2}>0\] \[x ^{2}(x ^{2}-1)>0\] \[x ^{2}>0 or x ^{2}>1\] x>0 or x<-1 or x>1 then x<-1 and x>0 will be the interval...

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klimenkov
  • klimenkov
\(x^2(x^2-1)>0\) divide it by \(x^2\): \((x-1)(x+1)>0\) \(x\in(-\infty;-1)\bigcup(1;\infty)\)
anonymous
  • anonymous
why is it greater than 0
anonymous
  • anonymous
oh subtract the x^2?
anonymous
  • anonymous
yep//....
anonymous
  • anonymous
@znimon understood the process?
anonymous
  • anonymous
almost, what about after x^2(x^2-1) is greater than 0
klimenkov
  • klimenkov
\(x^2\) is always bigger than 0. It equals 0 when \(x=0\). But it doesn't satisfy the inequality.
anonymous
  • anonymous
took x^2 common in both the terms....
anonymous
  • anonymous
yep got that
anonymous
  • anonymous
nops.. while splitting the inequality x has to be >0
anonymous
  • anonymous
will you break it down after the x^2(x^2-1) is greater than 0
anonymous
  • anonymous
yep.... both the split terms are compared to the inequalities...
anonymous
  • anonymous
I understand how it works so I need to find all the solutions for x that satisfy the inequality
anonymous
  • anonymous
thanks now I'll try some on my own. Will you recommend a problem that has the same properties?
anonymous
  • anonymous
@sriramkumar
anonymous
  • anonymous
try this one... \[x ^{8}-x ^{2}<0\] remeber x^3 has different properties when compared with x^2 .... have a good time!!!!1
anonymous
  • anonymous
thanks
anonymous
  • anonymous
@znimon you're welcome :)))))))))))))))))
anonymous
  • anonymous
\[\ \ x^4\ge x^2\\ \ \ x^2[x^2]\ge x^2\\x^2\text{ is either 0 non-zero.}\\ \text{if }x^2\text{ is 0,}\\ \ \ 0\times0=0\\ \ \ 0 = 0\\ \text{therefore }x^2[x^2]\ge x^2\\ \text{otherwise,}\\ \ \ x^2\ge1\\ \ \ |x|\ge\sqrt1\]
anonymous
  • anonymous
is either 0 or non-zero **
anonymous
  • anonymous
solution is \[x <0, x <1] so (negative infinity, to 0 not including zero)?
anonymous
  • anonymous
for x^3-x^2 is less than 0

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